My physics text asks "At what depth below the surface of a lake will you experience a pressure twice the pressure you experience on the surface?"

 

I'm guess if at sea level, the amount of pressure in atm exerted on me is exactly 1 atm.

 

So if its at 2 atm, there should be 2(760) torrs as well? So 1520 torrs = 1560mm Hg? Where do I go from here, I guess I can't use pV=nRT because it only applies to ideal GASES

[math]\rho gh[/math]

 

Remember that the atmospheric pressure is still present.

take a look at this site. http://www.engineersedge.com/fluid_flow/relationship_depth_pressure.htm

 

roughly, for every 33 ft of water you decend, there is an additional 1 atm of pressure (plus 1 for the atmosphere above the water)

edt

[math]\rho gh[/math]

 

Remember that the atmospheric pressure is still present.

 

The rule of thumb used in fluid mechanics is that the pressure experienced underwater at some depth h, is equal to the weight of the water directly above you. So this formula is correct.

yes but the fluid is already under pressure from the atmosphere unless you are talking about gauge pressure in which the atmosphere is defined as 0